Members Achilles Posted May 12, 2020 Members Posted May 12, 2020 (edited) Operatii pe numere mari In continuare voi prezenta cum se pot realiza operatii pe numere mari cu foarte putine linii de cod. In general, multi programatori se complica la aceste operatii, desi nu este nevoie! Vom considera ca numerele mari sunt vectori in care elementul de indice 0 indica lungimea numarului, iar cifrele sunt retinute in ordinea inversa decat cea a citirii. void add(int A[], int B[]) { int i, t = 0; for (i=1; i<=A[0] || i<=B[0] || t; i++, t/=10) A = (t += A + B) % 10; A[0] = i - 1; } Inmultirea unui numar mare cu un numar mic: void mul(int A[], int B) { int i, t = 0; for (i = 1; i <= A[0] || t; i++, t /= 10) A = (t += A * B) % 10; A[0] = i - 1; } Inmultirea unui numar mare cu un numar mare: void mul(int A[], int B[]) { int i, j, t, C[NR_CIFRE]; memset(C, 0, sizeof(C)); for (i = 1; i <= A[0]; i++) { for (t=0, j=1; j <= B[0] || t; j++, t/=10) C[i+j-1]=(t+=C[i+j-1]+A*B[j])%10; if (i + j - 2 > C[0]) C[0] = i + j - 2; } memcpy(A, C, sizeof(C)); } Scaderea a doua numere mari: void sub(int A[], int B[]) { int i, t = 0; for (i = 1; i <= A[0]; i++) { A -= ((i <= B[0]) ? B : 0) + t; A += (t = A < 0) * 10; } for (; A[0] > 1 && !A[A[0]]; A[0]--); } Impartirea unui numar mare la un numar mic: void div(int A[], int B) { int i, t = 0; for (i = A[0]; i > 0; i--, t %= B) A = (t = t * 10 + A) / B; for (; A[0] > 1 && !A[A[0]]; A[0]--); } Restul unui numar mare la un numar mic: int mod(int A[], int B) { int i, t = 0; for (i = A[0]; i > 0; i--) t = (t * 10 + A) % B; return t; } Sursa: Click Edited May 12, 2020 by Achilles
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